MHB No problem! Happy to help. (Smile)

[mathmari]

Hey! :giggle:Let $X_1, \ldots , X_n$ be independent, identically distributed random variables with $$P(X_i=-1)=P(X_i=1)=\frac{1}{2}$$
We consider the random variables $Y_i=\max \{X_i,X_{i+1}\}$, $i=1,\ldots , n-1$.
(a) Determine the distribution of $Y_i$, $i=1,\ldots , n-1$.
(b) Calculate the expected value of $Y_i$, $i=1,\ldots , n-1$.
(c) Calculate the covariance of $Y_i$ and $Y_j$, i.e. $\text{Cov}(Y_i, Y_j)=E(Y_iY_j)-E(Y_i)E(Y_j)$, $i,j=1,\ldots , n-1$.For (a) we have :
The results for $(X_i, X_{i+1})$ each with probability $\frac{1}{4}$ are :
\begin{equation*}(-1,-1), (1,-1), (-1,1), (1,1)\end{equation*}
When we consider of the two values eeach time we get $1$ in three cases and $-1$ in one case.
So we get \begin{equation*}P(Y_i = 1) = \frac{3}{4}, \ P(Y_i=-1) = \frac{1}{4}\end{equation*}

For (b)
The expected value of $Y_i$ is \begin{equation*}E[Y_i]=1\cdot P(Y_i = 1)+(-1)\cdot P(Y_i=-1)=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\end{equation*}

For (c) :
The covariance of $Y_i$ and $Y_j$ is \begin{equation*}\text{Cov}(Y_i, Y_j)=E(Y_iY_j)-E(Y_i)E(Y_j)=E(Y_iY_j)-\frac{1}{2}\cdot \frac{1}{2}=E(Y_iY_j)-\frac{1}{4}\end{equation*}
When $j>i+1$ then $Y_i$ and $Y_j$ are independent and so the covariance is $0$.
How can we calculate the covariance when $j=i+1$ ?:unsure:

 

[I like Serena]

For (c) :
The covariance of $Y_i$ and $Y_j$ is \begin{equation*}\text{Cov}(Y_i, Y_j)=E(Y_iY_j)-E(Y_i)E(Y_j)=E(Y_iY_j)-\frac{1}{2}\cdot \frac{1}{2}=E(Y_iY_j)-\frac{1}{4}\end{equation*}
When $j>i+1$ then $Y_i$ and $Y_j$ are independent and so the covariance is $0$.
How can we calculate the covariance when $j=i+1$ ?

Hey mathmari!

We can substitute the definitions of $Y_i$ and $Y_j$ and expand to the possible cases, just like we did in (a), can't we?

 

[mathmari]

We can substitute the definitions of $Y_i$ and $Y_j$ and expand to the possible cases, just like we did in (a), can't we?

We have that $Y_i=\max \{X_i, X_{i+1}\}$ and $Y_{i+1}+\max \{X_{i+1}, X_{i+2}\}$.

We have the possible values:
$(X_i, X_{i+1},X_{i+2})\in \{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1),(1,1,1) \} $

Right? What do we have to calculate now using this information? :unsure:

 

[I like Serena]

We can find the possible combinations of $Y_i$ and $Y_j$ and their probabilities, can't we?

 

[mathmari]

We can find the possible combinations of $Y_i$ and $Y_j$ and their probabilities, can't we?

Do we have to calculate first the product of $Y_i$ and $Y_j$? :unsure:

 

[mathmari]

Do we have to calculate first the product of $Y_i$ and $Y_j$? :unsure:

We have
\begin{equation*}(X_i, X_{i+1},X_{i+2})\in \{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1),(1,1,1) \}\end{equation*}
Then \begin{equation*}Y_iY_{i+1}\in \{(-1)\cdot (-1),(-1)\cdot 1,1\cdot 1,1\cdot (-1),1\cdot 1,1\cdot 1,1\cdot 1,1\cdot 1\}=\{1,-1,1,-1,1,1,1,1\}\end{equation*}
So $P(Y_iY_{i+1}=-1)=\frac{2}{8}$ and $P(Y_iY_{i+1}=1)=\frac{6}{8}$.

The expected value of $Y_iY_{i+1}$ is \begin{equation*}E[Y_iY_{i+1}]=1\cdot P(Y_i = 1)+(-1)\cdot P(Y_i=-1)=\frac{6}{8}-\frac{2}{8}=\frac{4}{8}=\frac{1}{2}\end{equation*}
So the covariance of $Y_i$ and $Y_j$ is \begin{equation*}\text{Cov}(Y_i, Y_j)=E(Y_iY_j)-\frac{1}{4}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\end{equation*}

:unsure:

 

[I like Serena]

Correct - if $j=i+1$. (Nod)

 

[mathmari]

Correct - if $j=i+1$. (Nod)

In other case it is equl to $0$, right? :unsure:

 

[I like Serena]

In other case it is equl to $0$, right?

Yes. (Nod)

 

[mathmari]

Yes. (Nod)

Great! Thank you! (Sun)